Algebra Flashcards
Definitions :
constant
term
coefficient
constant : a nb, or a symbol, such as Pi, that doesn’t change in value
term : a product of constants and variables, including powers of variables 5 x 6y² x²y²z²
coefficient : the constant factor of a term
in 6y², 6 is the coefficient
when no constant is written, the coefficient is 1
(a+b)² =
(a-b)² =
a²-b² =
(a+b)² = a² + 2ab + b² (a-b)² = a² - 2ab + b² a²-b² = (a+b)(a-b)
Factor the following expressions x² - 49 = 9x² -16 = 25x² - 64y² = x²y² - 1 =
x² - 49 = ( x + 7 )( x - 7 )
9x² -16 = ( 3x - 4 )( 3x + 4 )
25x² - 64y² = ( 5x + 8y )( 5x - 8y )
x²y² - 1 = ( xy - 1 )( xy + 1 )
Factor the following expressions x^6 - 16 = x^8 - 9y² = x^7 - 4x^5 = x^4 - 81 =
x^6 - 16 = ( x^3 - 4 ) ( x^3 + 4 )
x^8 - 9y² = ( x^4 + 3y ) ( x^4 - 9y)
x^7 - 4x^5 = x^5 ( x - 2 ) ( x + 2)
x^4 - 81 = ( x² + 9 ) ( x² - 9 ) = ( x² + 9 )( x - 3 )( x + 3 )
Factor the following expressions
y² - 25 =
25s² - t^4 =
x^9 - x =
y² - 25 = (y - 25)(y + 25)
25s² - t^4 = (5s - t²)(5s + t²)
x^9 - x = x(x^8 - 1) = x(x^4 - 1)(x^4 + 1) = x(x^4 + 1)(x² + 1)(x² - 1)
If y = 5 + x and y = 12 - x and if y² = x² + K then K equals which of the following? A : 17 B : 25 C : 60 D : 119
y = 5 + x so y - x = 5 y = 12 - x so y + x = 12
K = y² - x² = ( y - x )( y + x) = 5*12 = 60
(x + p)(x + q) =
x² + qx + px + pq
x² + (p + q)x + pq
we need to find the unknown constants p & q
the linear coefficient of the quadratic is the SUM of these two unknowns, and the constant term of the quadratic is the PRODUCT of the two unknowns
Factor : x² + 8x +15 x² + 14x + 24 x² + 4x -21 x² - 2x - 35 x² - 16x + 48
x² + 8x + 15
we need to find two numbers whose sum is 8 and whose product is 15
x² + 8x + 15 = (x+3)(x+5)
x² + 14x + 24 = (x+2)(x+12)
x² + 4x -21
since the sum is positive, the larger nb must be positive
+7 and -3
x² + 4x -21 = (x+7)(x-3)
x² - 2x - 35
one negative, one positive, one with bigger absolute value is negative
+5 and -7
x² - 2x - 35 = (x-7)(x+5)
x² - 16x + 48
both negative
-12 and -4
x² - 16x + 48 = (x-12)(x-4)
Factor : x² + 6x + 5 = x² - 5x -14 = x² + 6x -27 = x² - 20x + 36 =
x² + 6x + 5 = (x+5)(x+1)
x² - 5x -14 = (x+2)(x-7)
x² + 6x -27 = (x+9)(x-3)
x² - 20x + 36 = (x-2)(x-18)
Factor : 6x^3 - 150x = 2x² - 22x +48 = 7x^4 - 56x^3 - 63x² = -3x^9 + 48xy^4 =
6x^3 - 150x = 6x ( x² -25 ) = 6x (x-5) (x+5)
2x² - 22x +48 = 2 (x² - 11x +24) = 2 ( x - 8 ) ( x + 3 )
7x^4 - 56x^3 - 63x² = 7x² (x² - 8x - 9) = 7x² (x-9)(x+1)
-3x^9 + 48xy^4 = 3x (16y^4 - x^8) = 3x (4y² + x^4)(4y² - x^4) = 3x (4y² + x^4) (2y + x²) (2y - x²)
Find the prime factorization of 1599, 2491, 9975
notice that
1599 = 1600 - 1 = 40² - 1² = (40 + 1)(40 - 1) = (41)(39) =(41)(313)
413*13
2941 = 2500 - 9 = 50²- 3² = ( 50 -3)(50+3) = 4753
4753
9975 = 10 000 - 25 = 100² - 5² = (105)(95) = (521)(519) = (537)(519)
5²3719
0.9991 = 1 - 0.0009 = 1² - 0.03² = (1+0.03)(1-0.03) = (1.03)(0.97)
EXPRESS THE NB AS THE DIFFERENCE OF TWO SQARES
Simplify :
- 999951 / 0.993
(0. 999856/0.988) - 1
0.999951 / 0.993 = (1 - 0.000049) / (1 - 0.007)
= (1² - 0.007²) / (1 - 0.007)
= (1 + 0.007)(1 - 0.007) / (1 - 0.007)
= 1.007
(0.999856/0.988) - 1 = [(1 - 0.000144)/(1 - 0.012)] - 1
= [(1 - 0.012)(1 + 0.012) / (1 - 0.12)] - 1
= 1.012 - 1
= 0.012
Simplify :
( x² + 4x - 21) / ( x² - 6x + 9)
( 2x^4 - 8x² ) / ( x² - 5x - 14)
( y² + 2x - 8 ) / ( x - 4 )
( x² + 4x - 21) / ( x² - 6x + 9) = [(x+7)(x-3)] / (x-3)²
= (x+7) / (x-3)
( 2x^4 - 8x² ) / ( x² - 5x - 14) = [2x²(x² - 4)] / [(x+2)(x-7)]
= [2x² (x+2)(x-2)] / [(x+2)(x-7)]
= [2x²(x-2)] / (x-7)
( y² + 2x - 8 ) / ( x - 4 ) = [( y² / (x-4)] + [2(x-4)) / (x-4)]
= [y²/(x-4)] + 2
If (x² + 12x - 45) / ( x + 15) = 22, then x = ?
(x² + 12x - 45) / ( x + 15) = 22
(x+15)(x-3) / (x+15) = 22
x - 3 = 22
x = 25
If ( 4P + 3Q - 4R) / ( P - R ) = 19, then Q / (P - R) = ?
( 4P + 3Q - 4R) / ( P - R ) = 19 [4( P - R) + 3Q] / ( P - R ) = 19 4 + [3Q / (P-R)] = 19 3Q/(P-R) = 15 Q / (P - R) = 5
Simplify :
[ (x²/7) - (10x/7) + 3] / [ (x/6) - (1/2) ] = ?
multiply by the LCM of 6, 7 and 2 which is 42
42 [ (x²/7) - (10x/7) + 3] / 42 [ (x/6) - (1/2) ]
6(x² - 10x + 21) / 7(x - 3)
6(x-3)(x-7) / 7(x-3)
6(x-7) / 7
(6/7) * (x-7)
Solve :
x² + 13x - 40 = 8
3x² - 9x - 70 = 14
4x² - 4x - 4 = 14 + 8x - 2x²
x² + 13x - 40 = 8
x = -16 OR x = 3
3x² - 9x - 70 = 14
x = -4 OR x = 7
4x² - 4x - 4 = 14 + 8x - 2x²
x = -1 OR x = 3
Solve the system of equations :
A/ w - 2x + 3y = 13
B/ 2w + x - 4y = -14
C/ 3w - x + 2y = 8
pick two equations and eliminate one variable (substitution or elimination)
B + C = 5w - 2y = -6
5w = 2y - 6
pick another pair of equations and eliminate the same variable
A + 2B = 5w - 5y = -15
5w = 5y - 15
use the two equations with two unknowns technique
5y - 15 = 2y - 6
y = 3
5w = 2y - 6 w = 0
w - 2x + 3y = 13
x = -2
5x + 7 < 2x - 2
2/x > 1/3
-4 < 5 - 3x < 17
5x + 7 < 2x - 2
3x + 7 < -2
3x < -9
x < -3
2/x > 1/3
notice that x must unequal 0 and x > 0 because all nb are positive, we can cross multiply
6 > x
0 < x < 6
-4 < 5 - 3x < 17
-9 < -3x < 12
3 > x > -4
+ and - OK
x and / by positive OK
x and / by negative : reverse the direction of the inequality
What operations can we perform
if a < b and b < c ?
if a < b and c < d ?
if a < b and b < c :
we can combine inequalities in the same direction
a < b < c
if a < b and c < d :
we can add inequalities in the same direction
a + c < b + d
we can subtract inequalities in opposite directions
b - c < a - d
(no rule for multiplication or division)
Express |x - 7| < 3 as an ordinary inequality
we start at 7 on the nb line, and can go as far as a distance of 3 from that starting point
7 - 3 = 4
7 + 3 = 10
those are the endpoints, everything between them is allowed
4 < x < 10
Express the region -3 < x < 11 as an absolute value inequality
Find the middle of the region by averaging the endpoints
middle = 4
notice that x can be as far as 7 above 4 or 7 below 4
|x-4| < 7
Solve for x :
(2x - 1)² + 5(2x - 1) = 24
(2x - 1)² + 5(2x-1) - 24 = 0
Let u = 2x - 1 u² + 5u - 24 = 0 (u + 8)(u - 3) = 0 u + 8 = 0 OR u - 3 = 0 u = -8 OR u = 3
Use these values of u to solve for x
-8 = 2x - 1 OR 3 = 2x -1
x = -7/2 OR x = 2
Solve for x :
(x²+1)² - 15(x²+1) + 50 = 0
Let u = x²+1 u² - 15u + 50 = 0 (u - 5)(u - 10) = 0 u - 5 = 0 OR u - 10 = 0 u = 5 OR u = 10
x² + 1 = 5 OR x² + 1 = 10
x² = 4 OR x² = 9
x = +/- 2 OR x = +/- 3
Solve for k:
3 / [ 1 - 8/(7 + k)] = 15
Let A = 1 - 8/(7 + k)
Then 3/A = 15 so 3 = 15 A and A = 1/5
Let B = 8 / (7+k)
Then A = 1 - B so 1/5 = 1 - B and B = 1 - 1/5 = 4/5
Then 4/5 = 8 / (7+k)
so 1/5 = 2 / (7+k)
7 + k = 10
k = 3
